d) I'm not sure how to exploit the orthogonal properties, but I do know ##\int \psi^{*}\ \psi dx = 1##, So I chose one of the components of the wave function with the unknown constant ##A##, and did exactly this.

a) This point asks you to mention the properties of operators whose associated quantum numbers is 'good', and second a property that allows us to write a unique eigenstate defined by a collective numbers composed of each of the quantum numbers from the operators A, B, C, ...

b) But there is a reason why that identity can be true. For a simpler case with only one quantum number, use the fact that ##\langle a'|A| a \rangle##, where the ##|a \rangle##'s are eigenstates of operator A, can be calculated in two ways.

c) Remember that the modulus square of a certain expansion coefficient in the wavefunction represented in some basis is interpreted as a probability to find that wavefunction to be in that basis state. How will you use this information to calculate A?

d) A is supposed to be already computed in c). For this point, simply use the orthonormality of the eigenstates of H atom and the interpretation of the expansion coefficient as I have mentioned in c).

What about googling or reading in the book where the problem is? I'm almost sure, that problem is not the first place where the term 'good quantum number' appears. Just for hint, this is related with time evolution operator.

Which do you refer by 'they'?
Allright to get you further in this problem, I'll let you know the mathematical property which an operator A whose quantum number is said to begood must satisfy, that is [H,A] = 0. But the problem asks you to interpret the physical implication of that relation.

Yes, it must be conserved in time. If it's not, it will be inconvenient for us since if we allow our system to evolve in time, we must always compute the new quantum number for that particular time as our initial quantum number has changed. That's the reason why it's always desireable to seek for the good quantum numbers for a system before being able to represent any eigenstate as a unique collective indices of those quantum numbers.

I mean, for an arbitrary operator ##A##, then ##A^{2}## and ##A_{z}## have simultaneous eigenfunctions? I mean, I was using orbital angular momentum as a concrete example. Who is to say that there is a z-component of some arbitrary operator?

No, such analogy is unnecessary, you can only define spatial components of an operator if it is a vector operator that satisfies its own requirement.

For H atom the operators H, L2, Lz and Sz share a common set of eigenstate commonly written as ##|nlm_lm_s\rangle##. What property do all those 4 operators have in relation with certain mathematical relation?

I am not sure what you are trying to get me to say, is it that they have simultaneous eigenstates??

Imagine you have arbitrary state ##|\psi \rangle##, now you apply this operator: ##AB-BA## on to our arbitrary state where A (with quantum number ##a##) and B (with quantum number ##b##) are known to have a common set of eigenstates, which we write as ##|a,b \rangle##. Hence we have relations like ##A|a,b\rangle = a|a,b\rangle## and ##B|a,b\rangle = b|a,b\rangle## . Since ##|a,b \rangle##'s are eigenstates, they must be complete and therefore we can expand ##|\psi \rangle## into the bases ##|a,b \rangle## for all ##a,b##. My question is, given those forementioned conditions, what is the result of ##(AB-BA)|\psi \rangle##?